Convergence of Fourier series where function is continuous

Question

I'm not able to understand how they worked out for x not equal to 2*pi*n the series converges to x mod 2*pi

Any help would be much appreciated

Answer 1

Start with

$\sum\limits_{n = 1}^\infty {\frac{{\sin (nx )}}{n}} = \left\{ \begin{array}{l} \frac{1}{2}(\pi - x )\;,\quad 0 < x < 2\pi \\ 0\quad x = 0,\quad 2\pi \end{array} \right.$ .

Then $\pi - 2\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)}}{n}} = \left\{ \begin{array}{l} x\;,\quad 0 < x < 2\pi \\ \pi \quad x = 0,\quad 2\pi \end{array} \right.$ .

At $0$ and all multiples of $ 2\pi $, the value is $ \pi $ and by periodicity at all other values the value between $0$ and $2 \pi $ is repeated. Hence except for multiples of $2 \pi $, the value is the remainder after dividing by $2 \pi $. So we can say for values of $x$ not a multiple of $2 \pi $ the series is given by $x$ mod $2 \pi $ .

You can use complex power series to obtain the first summation above.